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The time between events in a Poisson process is described by the pdf $f(t; T_0)=\frac{1}{T_0}e^{-t/T_0}$.

I'm interested in estimating how long we would have to observe this process before we see two events separated by at least a specified time $T_{min}$. In other words, how long do I need to observe the process before I see an event, while not having seen an event in the previous time $T_{min}$?

Concrete example: the time between buses arriving at a stop is described by an exponential distribution with mean 10 minutes. How long do I have to wait at the bus stop before I see a bus, while not having seen a bus in the past 5 minutes?

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This answer requires solving three subproblems:

  1. How many bus arrivals $N$ do I expect to suffer through before experiencing a time between bus arrivals $> T_{min}$?

  2. How long is the expected wait between buses $\mathbb{E}T_s$ given that $T_s \leq T_{min}$?

  3. How long is the expected wait between two consecutive buses $\mathbb{E}T_l$ given that $T_l > T_{min}$?

The final answer is equal to $\mathbb{E}N \mathbb{E}T_s + \mathbb{E}T_l$, or perhaps just $\mathbb{E}N \mathbb{E}T_s$ if the time spent in the last interval isn't counted.

The answer to 3 is quite easy, thanks to the Exponential / Poisson assumption. The memoryless property of the Exponential distribution implies that $\mathbb{E}T_l = T_{min} + T_0$.

The answer to 2 is a little more involved; we have to find the expected value of the appropriate conditional random variate:

$$\frac{\int_0^{T_{\min}}xe^{-x/T_0}\text{d}x}{T_0(1-e^{-T_{min}/T_0})}$$

which, using integration by parts, solves to:

$$\frac{T_0 - (T_0+T_{min})e^{-T_{min}/T_0}}{1-e^{-T_{min}/T_0}}$$

As @Moormanly has observed, $N$ is distributed Geometric, in this case with parameter $p = \exp\{-T_{min}/T_0\}$. The expected value of a Geometric$(p)$ distribution is $(1-p)/p$, so, substituting and writing the full equation out, we get:

$$\left(\frac{1-e^{-T_{min}/T_0}}{e^{-T_{min}/T_0}}\right)\left(\frac{T_0 - (T_0+T_{min})e^{-T_{min}/T_0}}{1-e^{-T_{min}/T_0}}\right)+T_{min}+T_0$$

where the last two terms are optional depending upon the exact nature of the problem being solved. Making the obvious cancellation leads to:

$$\frac{T_0 - (T_0+T_{min})e^{-T_{min}/T_0}}{e^{-T_{min}/T_0}}+T_{min}+T_0$$

and some further algebra gets us to:

$$\frac{T_0}{e^{-T_{min}/T_0}}$$

Checking our results via simulation on your sample problem comes next. $T_0 = 10$ and $T_{min}=5$ gives us a calculated result of 16.483. Our simulation code:

res <- rep(0,10000)
for (i in 1:length(res)) {
   res[i] <- x <- rexp(1,1/10)
   while (x < 5) {
      x <- rexp(1,1/10)
      res[i] <- res[i] + x
   }
}

c(mean(res), (mean(res)-10/exp(-0.5))/(sd(res)/sqrt(length(res))))

reports the simulated mean and the associated t-statistic for testing whether or not it is equal to the calculated value. The results are:

> c(mean(res), (mean(res)-10/exp(-0.5))/(sd(res)/sqrt(length(res))))
[1] 16.5793553  0.8998745

which would seem to confirm that we haven't messed up our algebra anywhere.

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Think first about the number of events you expect to observe, think next about the amount of time it will take to observe those events.

Letting $N$ be the number of events before an inter-arrival time of at least $T_{min}$, see that $N$ is geometric with parameter $p=P(T_i>T_{min})$. Now let $T$ be the amount of time taken to observe these events. Expanding using total expectation: $$E[T]\\=\sum_{n=1}^\infty P(N=n) E[T|N=n]\\=\sum_{n=1}^\infty P(N=n) \bigg((n-1)E[T_i|T_i<T_{min}]+E[T_i|T_i \ge T_{min}]\bigg) $$

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